Answer:
It is [tex]74.24[/tex] degrees Fahrenheit when the crickets are chirping [tex]150[/tex] times a minute.
Variables [tex]x = Chirps [/tex] and [tex]y = Degrees[/tex]
y-intercept [tex]= 38.24[/tex]
slope [tex]= 0.24[/tex]
Step-by-step explanation:
As the chirps are [tex]124[/tex] times per minute for [tex]68[/tex] degrees Fahrenheit we will consider it in [tex](x_1,y_1) = (124,68)[/tex].
Now when the chirps are [tex]172[/tex] times and temperature is [tex]80[/tex] degrees we consider it [tex](x_2,y_2) = (172,80)[/tex].
To find the slope we can use point-slope formula.
Where [tex]m[/tex] is the slope and [tex]m=\frac{y_2-y_1}{x_2-x_1} =\frac{80-68}{172-124}= 0.24[/tex]
To find y-intercept we will plug out the value of slope in [tex]y=m(x)+b[/tex] choosing point [tex](x_1,y_1) = (124,68)[/tex].
So [tex]b=y-m(x) =68-124(0.24) =68-29.76=38.24[/tex]
And the equation is [tex]y=0.24(x)+b[/tex]
To calculate temperature for [tex]150[/tex] chirpings we have to plug [tex]x=150[/tex] and [tex]b=38.24[/tex] in our above equation which is in form of [tex]y=m(x)+b[/tex]
[tex]y=0.24(x)+b[/tex]
[tex]y=0.24(150)+38.24=74.24\approx 75[/tex] degrees Fahrenheit.
Finally we have :
Temperature [tex]74.24[/tex] degrees Fahrenheit.
With variables [tex]x = Chirps [/tex] and [tex]y = Degrees[/tex]
y-intercept [tex]= 38.24[/tex] and
slope [tex]= 0.24[/tex]
